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a) \(\left|2x+1\right|=\left|1-x\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=1-x\\2x+1=x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
b) \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
c) \(\left|2x-3\right|-\left|3x+2\right|=0\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\5x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)
d) \(\left|2+3\right|=\left|4x-3\right|\Leftrightarrow\left|4x-3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}4x-3=5\\4x-3=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=8\\4x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
e) \(\left|\frac{5}{4}-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\Leftrightarrow\left|\frac{5}{8}x+\frac{3}{5}\right|=\frac{9}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{3}{5}=\frac{9}{4}\\\frac{5}{8}x+\frac{3}{5}=-\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{33}{20}\\\frac{5}{8}x=-\frac{57}{20}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{66}{25}\\x=-\frac{114}{25}\end{cases}}\)
\(\left|2x+1\right|=\left|1-x\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=-x+1\\2x+1=x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+1\\2x-x=-1-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
b. \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=4+2\\5x+x=4-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
c. \(\left|2x-3\right|-\left|3x+2\right|=0\)
\(\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=3+2\\2x+3x=3-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=5\\5x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)
d, e tương tự
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
c) x.(1+2+3+4+...+100)=0
x.5050=0
x=0:5050=0
Vậy x=0
d) x.(1+2+3+4+5+...+100)=5050
x.5050=5050
x=1
Vậy x=1
e) x+1+x+2+x+3+x+4+...+x+100=5050
(x+x+x+x+...+x)+(1+2+3+4+...+100)=5050
100 số hạng x
x.100+5050=5050
x.100=0
x=0
Vậy x=0
a,x.(x+7)=0
suy ra x=o hoặc x+7=0
vs x+7=0
x=0+7
x=7
vậy x=0 hoặc x=7
b(2+2x)(7-x)=0
suy ra 2+2x=0 hoặc 7-x=0
vs2+2x=0 vs7-x=0
2x =0-2 x=0+7
2x =(-2) x=7
x=(-2);2
x=-1
vậy x=-1 hoặc x=7
d(x^2-9)(3x+15)=0
suy ra x^2-9=0 hoặc 3x+15=0
vsx^2-9=0 vs 3x+15=0
x^2 =0+9 3x =0-15
x^2 =9 3x =-15
x^2 =3^2 x=(-15):3
suy ra x=3 hoặc x=-3 x=-5
vậy x=3 x=-3 hoặc x=-5
e,(4x-8)(x^2+1)=0
suy ra4x-8=0 hoặc x^2+1=0
vs 4x-8=0 vs x^2+1=0
4x =0+8 x^2 =0-1
4x =8 x^2 =-1
x =8:4 x^2 =-1^2 hoặc 1^2
x =2 suy ra x=-1 hoặc x=1
vậy x=2, x=-1 hoặc x=1
a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x-7\\x+2=-3x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-2-7\\x+3x=-2+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Mấy câu kia tương tự.
a) \(\left(x+2\right)^2-\left(3x-7\right)^2=0\)
\(\Leftrightarrow\left(x+2-3x+7\right)\left(x+2+3x-7\right)=0\)
\(\Leftrightarrow\left(-2x+9\right)\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+9=0\\4x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9}{-2}=\dfrac{9}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(x=\dfrac{9}{2}\) hoặc \(x=\dfrac{5}{4}\)
b) lộn đề à
c) \(25\left(x-3\right)^2-49\left(2x+1\right)^2=0\)
\(\Leftrightarrow5^2\left(x-3\right)^2-7^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[5\left(x-3\right)\right]^2-\left[7\left(2x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(5x-15\right)^2-\left(14x+7\right)^2=0\)
\(\Leftrightarrow\left(5x-15-14x-7\right)\left(5x-15+14x+7\right)=0\)
\(\Leftrightarrow\left(-9x-22\right)\left(19x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x-22=0\\19x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x=22\\19x=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{22}{-9}=\dfrac{-22}{9}\\x=\dfrac{8}{19}\end{matrix}\right.\)
Vậy \(x=\dfrac{-22}{9}\) hoặc \(x=\dfrac{8}{19}\)
d) \(9\left(3x-2\right)^2=121\left(1-4x\right)^2\)
\(\Leftrightarrow9\left(3x-2\right)^2-121\left(1-4x\right)^2=0\)
\(\Leftrightarrow3^2\left(3x-2\right)^2-11^2\left(1-4x\right)^2=0\)
\(\Leftrightarrow\left[3\left(3x-2\right)\right]^2-\left[11\left(1-4x\right)\right]^2=0\)
\(\Leftrightarrow\left(9x-6\right)^2-\left(11-44x\right)^2=0\)
\(\Leftrightarrow\left(9x-6-11+44x\right)\left(9x-6+11-44x\right)=0\)
\(\Leftrightarrow\left(53x-17\right)\left(-35x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}53x-17=0\\-35x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}53x=17\\-35x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{53}\\x=\dfrac{-5}{-35}=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(x=\dfrac{17}{53}\) hoặc \(x=\dfrac{1}{7}\)